a single frictionless roller coaster car of mass m=750 kg tops the first hill with speed v=15 m/s ar height h =40m as shown

1) Find the speed of the car at B and C

2) if mass m were doubled, would the speed at B increase, decrease, or remain the same?

¡Hellow!

How we know, for calculate density, we can use the formule:

                                               [tex]\boxed{\textbf{p = m / v} }[/tex]

Where:

[tex]\sqrt{[/tex] p = Density = ?

[tex]\sqrt{[/tex] m = Mass = 57,9 g

[tex]\sqrt{[/tex] v = Volume = 3 cm³

Let's replace according we information:

p = 57,9 g / 3 cm³

p = 19,3 g/cm³

The density of the metal bar is 19,3 g/cm³.

¿Good luck?

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

where

m is the mass

f is the force

We have

[tex]a = \frac{275}{18} \\ = 15.277777...[/tex]

We have the final answer as

Hope this helps you

Answer:

Explanation:

in this problem, we are expected to solve for the maximum height of a projectile (the ball)

Firstly, the expression for maximum height is

h = u ^2 *sin ^2 θ /2 g

Given data

initial velocity u=22m/s

angle of projection, θ= 40degrees

substituting into the expression we have

h = 22^2 *sin (40)^2 /2 *9.81

h = 484*0.413 /19.62

h =199.892 /19.62

h=10.18m

The maximum height is 10.18m

The maximum height attains by ball is 48.64 meter.

The maximum height is given by using formula,

                  [tex]h=\frac{u^{2}*sin2\theta }{g}[/tex]

Where u is initial velocity, and value of [tex]g=9.8m/s^{2}[/tex]

Given that,   [tex]u=22m/s,\theta=40,g=9.8m/s^{2}[/tex]

Substituting above values in formula.

             [tex]h=\frac{(22)^{2}*sin80 }{9.8} =\frac{484*0.98}{9.8} =48.64m[/tex]

Hence, the maximum height attains by ball is 48.64 meter.

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The kinetic energy of a body determines it's level of impact on the object in which it comes in contact with. Hence, the much larger kinetic energy exhibited by a wrecking ball compared to a yo-yo means that is has a much larger impact on a building than a yo-yo.

The kinetic energy of a body is a factor of it's velocity and mass as they are directly proportional.

The wrecking ball has a very large mass which is thousands of times larger than that of a yo-yo. Also, the velocity at which a wrecking ball is launched is higher than the velocity of a yo-yo.

This means that the kinetic energy of a wrecking ball is much higher than that of yo-yo. Hence, having much greater impact on a building compared to a yo-yo.

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Answer:

answer: 105 sin 30 = 52.5

Which object? More information is needed to answer this question

If the free ends of conduction tester are connected for long time , The cell of the battery will drain very quickly

Answer:

The table is 2.08 m high

Explanation:

Horizontal Motion

When an object is thrown horizontally with a speed vo from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:

[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]

If we know the range d = 0.15 m and the speed v = 0.23 m/s, we can solve the above equation for h:

[tex]\displaystyle h=\frac{d^2\cdot g}{2v^2}[/tex]

[tex]\displaystyle h=\frac{0.15^2\cdot 9.8}{2\cdot 0.23^2}[/tex]

[tex]\boxed{h=2.08\ m}[/tex]

The table is 2.08 m high

Answer:

[tex]h_p = 30.46\ m[/tex]

Explanation:

Free Fall Motion

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

[tex]v_f=g\cdot t[/tex]

Where [tex]g = 9.8 m/s^2[/tex]

Similarly, the distance y the object has traveled is calculated as follows:

[tex]\displaystyle y=\frac{g\cdot t^2}{2}[/tex]

If we know the height h from which the object was dropped, we can solve the above equation for t:

[tex]\displaystyle t=\sqrt{\frac{2\cdot y}{g}}[/tex]

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

[tex]\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec[/tex]

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

[tex]t_2=2.56 - 2 = 0.56\ sec[/tex]

Therefore, it has traveled down a distance:

[tex]\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m[/tex]

Thus, the height of the pen is:

[tex]h_p = 32 - 1.54\Rightarrow h_p=30.46\ m[/tex]

The pen is 30.52 m above the ground.

Given that the height of the stadium is h = 32m

The initial velocity of the glasses will be 0.

[tex]h=\frac{1}{2}gt^{2} \\t=\sqrt{\frac{2h}{g} } \\t=\sqrt{\frac{2*32}{9.8} }\\t=2.55s[/tex]is the time taken for the glasses to hit the ground.

Now the pen is released 2 seconds later. So by the time the glasses hit the ground the pen has spent:

[tex]t^{'}=2.55-2\\t^{'}=0.55s[/tex]in the air

distance traveled by the pen:

[tex]d=\frac{1}{2}gt^{2}\\\\d=\frac{1}{2}*9.8*0.55*0.55\\\\d=1.48m[/tex]

So the pen is  [tex]h-d=32-1.48=30.52m[/tex]  above the ground.

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Answer:

an electro magnet is technology contain at least one permanent magnet

Answer:

The answer is D, every object in the universe attracts every other objects

Explanation:

Newton's law of universal gravitation is described as every object in the universe attracts every other objects.

Gravitation is the force in physics that draws two masses together. Unbelievably, every single particle of matter in the cosmos attracts every other particle through gravity. The phrases gravitation and gravity are sometimes used synonymously to refer to the attraction that exists between everything with mass or energy.

According to Newton's Law of Universal Gravitation, every particle in the cosmos is drawn to every other particle with a force that is directly proportional to the product of their masses and inversely proportional to their distance from one another.

So, every object in the universe attracts every other objects describes  Newton's law of universal gravitation.

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#SPJ6

Answer:

Atoms in a solid are close and tightly packed.They vibrate about fixed positions .They are arranged in a regular LATTICE.They cannot be compressed nor can they change shape.They have strong forces of attraction between them so they cannot move in order to move they have to gain ENERGY. Whereas the atoms in a liquid are not that close compared to the ones in solid.They take the shape of the container but their volume remains fixed.They have weak forces of attraction between atoms and have more energy which enables them to move.

Explanation:

Answer:

0.5 m/s²

Explanation:

Net force = mass × acceleration

∑F = ma

20,000 N = (40,000 kg) a

a = 0.5 m/s²

Answer:

which one will not start moving faster

Explanation:

Mass is also described by physics. If you apply the same amount of force to a small sports car and to a big garbage truck.

Answer:

The second choice

Explanation:

I think the answer is the second choice because if the surface is smooth, there is less friction. With a boat, it is easier to pull it on water than on the sand, because water has less friction, and thus, the answer is the second choice because rough surfaces have more friction.

Hopefully the explanation and answer helps!

Answer:

The new freezing point of water is -0.476°C.

Explanation:

Given that,

Mass of water = 4.62 kg

Mass of CaBr₂ = 236.5 g

We need to calculate the molality of solution

Using formula of molality

[tex]m=\dfrac{mass\ of\ CaBr_{2}}{molar\ mass\ of\ CaBr_{2}\times mass\ of\ water}[/tex]

Put the value into the formula

[tex]m=\dfrac{236.5}{200\times4.62}[/tex]

[tex]m=0.256\ mol/kg[/tex]

[tex]m=0.256\ m[/tex]

We need to calculate the depression in freezing point of water

Using formula of freezing point

[tex]\Delta T=m\times k[/tex]

Where, k = depression constant

m = molality of solution

Put the value into the formula

[tex]\Delta T=0.256\times1.86[/tex]

[tex]\Delta T=0.476^{\circ}C[/tex]

We know that,

The standard freezing point of water is 0°C

We need to find the new freezing point of water

Using formula for freezing point of water

Freezing point of 4.62 kg of water = freezing point of water-depression in freezing point of water

[tex]\Delta T'=0-\Delta T[/tex]

Put the value into the formula

[tex]\Delta T'=0-0.476[/tex]

Hence, The new freezing point of water is -0.476°C.

Answer:

equal, opposite

Explanation:

Newton’s third law says that every time there is an interaction force, there is also a force that is equal in size and acts in the opposite direction .

Answer:

  about -1.26 m/s²

Explanation:

The acceleration can be found from the formula ...

  a = (v2² -v1²)/(2d)

Here, we have the initial velocity v1=23 m/s, the final velocity v2=0, and teh distance d=210 m. Using the formula, we find the acceleration to be ...

 a = (0 -23²)/2(210) ≈ -1.2595 . . . m/s²

Answer:

A

Explanation:

Switches are used to block electrical currents, thus needing to be made out of insulating material.

Answer:I think the answer is 5.43 I don’t rlly know

Explanation:

Answer:

D

Explanation:

Cells are the smallest unit of organizaiton, followed by tissues, organs, organ systems, and then organisms themselves.

Answer:

20.83rad/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the mass of the disc and Brick respectively

u1 and u2 are the linear velocities respectively

v is their final velocity of the block-disc system

Given

m1 = 100kg

m2 = 20kg

u1 = wr =25×1.6

u1 = 40m/s

u2 = 0m/s (brick is stationary)

Substitute and the final linear velocity v

100(40)+20(0) = (100+20)v

4000 = 120v

v = 4000/120

v = 33.3m/s

From the formula;

v = wr

w = v/r

w= 33.3/1.6

w = 20.83rad/s.

Hence the final angular speed of the disc-brick system is 20.83rad/s

Hellow!

First, lets recabe information:

m (mass) = 1500 kg

a (aceleration) = 3 m/s²

F (Force) = ?

For calculate Force, if we have mass and aceleration, we can applicate the second law of Newton:

"Force is proportional to the product of the mass plus aceleration."

In a equation, this is like:

F = m * a

Let's replace according we information, and let's resolve:

F = 1500 kg * 3 m/s²

F = 4500 N

Result:

The force applicated is of 4500 Newtons.

Gud Luck!

Answer:

1- bounce rule

2- left

3-11,2

4-true

5- whiffle ball

6- kitchen

7-false

8- ping pong

9- true

10- true

Explanation:

I'm pre5sure they are all right. brainliest please

Answer: I had the same quiz too I’m only doing so the other guy can get brainliest

Explanation:Kfkfkfjgjggj

9514 1404 393

Answer:

  0.2 m/s

Explanation:

The average velocity is the change in position divided by the change in time:

  (8 m -4 m)/(20 s) = (4/20) m/s = 1/5 m/s = 0.2 m/s

The boy's average velocity is 0.2 m/s.

Answer:

44.3 m/s

Explanation:

a) Draw a free body diagram of the mass M.  There are three forces:

Weight force mg pulling down,

Normal force N pushing perpendicular to the ramp,

and tension force T pulling parallel up the ramp.

Sum of forces in the parallel direction:

∑F = ma

T − Mg sin 30° = 0

T = Mg sin 30°

T = Mg / 2

Draw a free body diagram of the hanging mass m.  There are two forces:

Weight force mg pulling down,

and tension force T pulling up.

Sum of forces in the vertical direction:

∑F = ma

T − mg = 0

T = mg

Substitute:

mg = Mg / 2

m = M / 2

M = 2m

b) Velocity of a standing wave in a string is:

v = √(T / μ)

T = mg, and m = 5 kg, so T = (5 kg) (9.8 m/s²) = 49 N.  Therefore:

v = √(49 N / 0.025 kg/m)

v = 44.3 m/s

Answer:

it will take half of the initial time to reach double the velocity

Explanation:

Given that a car accelerates from rest to a certain velocity in a certain time.

Since the car accelerate from rest, the initial velocity U = 0

Using the first equation of motion,

V = U + at

V = at

Assume that there is no friction, and that the engine power is constant. Consider the following scenarios independently. How long would it take to reach double the velocity (also from rest)

That is V = 2V

Using the same formula ,

2( at ) = at

2T = t

T = t/2

Therefore, it will take half of the initial time to reach double the velocity

Answer:

2.5m/s^2

Explanation:

Note that the boat is reducing its speed. It is having negative acceleration or deceleration.

V^2 = u^2 -2as ( minus sign is used because of speed is reduced)

Given that,

s = 12 m

v = 5.5 m/s

u = 9.5 m/s

a = (v^2 - u^2) ÷ (-2s)

a= ( 5.5^2 - 9.5^2) ÷ ( -2× 12)

a = 2.5 m/s^2

The acceleration of the boat is -2.5 m/s²

The given paramters;

distance traveled by the boat, d = 12 m

initial velocity of the boat, u = 9.5 m/s

final velocity of the boat, v = 5.5 m/s

The acceleration of the boat is calculated as;

[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(5.5)^2 -(9.5)^2 }{2(12)} \\\\a = -2.5 \ m/s^2[/tex]

Thus, the deceleration of the boat is 2.5 m/s²

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